Problem: What is the sum of the first 10 terms of the arithmetic sequence that begins $-12, -8, -4,...$?
The sum of the first $n$ terms of an arithmetic series is: $\frac{n(a_1 + a_n)}{2}$, where $a_1$ is the first term of the sequence and $a_n$ is the last. $a_n = a_1 + (n - 1)d$, where $d$ is the difference between consecutive terms. For this specific sequence, $a_1 = -12$ and $d = 4$. Thus, the sum of the first 10 terms of the sequence is: $\frac{10(a_1 + a_{10})}{2} = 5(a_1 + a_1 + (n - 1)d) = 5((-12) \cdot 2 + 9 \cdot 4) = 5 \cdot (-24 + 36) = 5 \cdot 12 = \fbox{60}$.